Three nonconducting infinite sheets are parallel to the yz plane of an xyz coord

Question

Three nonconducting infinite sheets are parallel to the yz plane of an xyz coordinate system. Each sheet has a uniform surface charge density. Sheet 1, negatively charged with surface charge density , passes through the x axis at x=1.0 = 1.0 m . Sheet 2 has an unknown surface charge density and passes through the x axis at x = 2.0 m . Sheet 3, negatively charged with surface charge density 3, and passes through the x axis at x = 4.0 m . The electric field due to the sheets is zero at x = 1.5 m .

A) What is the surface charge density on sheet 2?

B) If the electric field at x = 0 is E0, what is the electric field magnitude at x = -2.0 m ?

C) If the electric field at x = 0 is E0, what is the electric field magnitude at x = 3.0 m ?

D) If the electric field at x = 0 is E0, what is the electric field magnitude at x = 6.0 m ?

Explanation / Answer

as we know, electric field due to an infinite sheet of charge is given by

E=pho/(2*epsilon)

where pho=surface charge density

epsilon=electrical permitivity of the medium(in this case, free space)=8.85*10^(-12)

direction of field is away from the sheet and towards the point if charge density is positive

and away from the point and towards the sheet if charge density is negative

Q(A).

at x=1.5 m, field due to sheet 1 will be towards the sheet 1(as sheet 1 carries negative charge)

and hence towards -ve x axis.

field magnitude=sigma/(2*epsilon)

at x=1.5 m, field due to sheet 3 will be towards the sheet 3(as sheet 3 carries negative charge)

and hence towards +ve x axis.

field magnitude=3*sigma/(2*epsilon)

hence total field along +ve x axis=2*sigma/(2*epsilon)

as it is given that total field at x=1.5 m is 0, the field due to sheet 2 will be along -ve x direction and of magnitude 2*sigma/(2*epsilon)

hence the surface charge density will be 2*sigma and positive.

part B:

field at origin:

field due to sheet 1 at origin will be towards sheet 1 and hence towards +ve x axis.

magnitude of field=sigma/(2*epsilon)

field due to sheet 2 at origin will be away from sheet 2 and hence towards -ve x axis.

magnitude=2*sigma/(2*epsilon)

field due to sheet 3 at origin will be towards sheet 3 and hence +ve x axis.

mangitude =3*sigma/(2*epsilon)

hence total field is along +ve x axis with magnitude 2*sigma/(2*epsilon)

hence E0=2*sigma/(2*epsilon) and direction is along +ve x axis.

field at x=-2:

field due to sheet 1 at x=-2 will be towards sheet 1 and hence towards +ve x axis.

magnitude of field=sigma/(2*epsilon)

field due to sheet 2 at x=-2 will be away from sheet 2 and hence towards -ve x axis.

magnitude=2*sigma/(2*epsilon)

field due to sheet 3 at x=-2 will be towards sheet 3 and hence +ve x axis.

mangitude =3*sigma/(2*epsilon)

hence total field is along +ve x axis with magnitude 2*sigma/(2*epsilon)

i.e. field at x=-2 m is equal to E0.

part c:

field due to sheet 1 at x=3 will be towards sheet 1 and hence towards -ve x axis.

magnitude of field=sigma/(2*epsilon)

field due to sheet 2 at x=3 will be away from sheet 2 and hence towards +ve x axis.

magnitude=2*sigma/(2*epsilon)

field due to sheet 3 at x=3 will be towards sheet 3 and hence +ve x axis.

mangitude =3*sigma/(2*epsilon)

total field will be along +ve x axis with magnitude=4*sigma/(2*epsilon)

hence field at x=3 m is 2*E0.

part D:

field due to sheet 1 at x=6 will be towards sheet 1 and hence towards -ve x axis.

magnitude of field=sigma/(2*epsilon)

field due to sheet 2 at x=6 will be away from sheet 2 and hence towards +ve x axis.

magnitude=2*sigma/(2*epsilon)

field due to sheet 3 at x=6 will be towards sheet 3 and hence -ve x axis.

mangitude =3*sigma/(2*epsilon)

total field is along -ve x axis and magnitude=2*sigma/(2*epsilon)

hence total field at x=6 m is -E0.

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